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Fundamental modes, harmonics and beats: NEET MCQs with answers

75 questions on this topic of Oscillations and Waves (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. A wire of length L and mass m, clamped at both ends under tension T, vibrates in its fundamental mode. If the tension is increased by 44% while keeping length and mass constant, the percentage increase in its fundamental frequency is:

  1. 12%
  2. 22%
  3. 44%
  4. 20%
Show answer and explanation

Answer: (D) 20%

The fundamental frequency of a stretched string is f = (1 / 2L) √(T / μ). If tension T' = 1.44 T, then the new frequency is f' = f √(1.44) = 1.2 f. Thus, the percentage increase is ((1.2 - 1) / 1) × 100% = 20%.

medium

2. A glass tube of length 1.0 m is held vertically and filled completely with water. A drain tap at the bottom is opened while a vibrating tuning fork of frequency 500 Hz is held over the top open end. If the speed of sound in air is 340 m/s, how many resonances can be observed as the tube drains completely? (Neglect end correction)

  1. 1
  2. 2
  3. 3
  4. 4
Show answer and explanation

Answer: (C) 3

Resonances occur at air column lengths l = (2n - 1)λ / 4. The wavelength is λ = v / f = 340 / 500 = 0.68 m = 68 cm. The successive resonance lengths are l₁ = 68 / 4 = 17 cm, l₂ = 3 × 17 = 51 cm, and l₃ = 5 × 17 = 85 cm. The next resonance would occur at 7 × 17 = 119 cm, which exceeds the tube length of 100 cm. Hence, exactly 3 resonances are observed.

hard

3. A stationary wave on a stretched string of length 1.2 m clamped at both ends is given by y(x, t) = 0.06 sin(5πx / 3) cos(250πt), where x and y are in meters and t is in seconds. The total number of nodes (including the clamped ends) and the speed of the constituent progressive waves are:

  1. 4 nodes, 150 m/s
  2. 3 nodes, 150 m/s
  3. 3 nodes, 300 m/s
  4. 4 nodes, 300 m/s
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Answer: (B) 3 nodes, 150 m/s

The wave number is k = 5π / 3 m⁻¹. The wavelength is λ = 2π / k = 2π / (5π / 3) = 6 / 5 = 1.2 m. Since the string length is L = 1.2 m = λ, the string is vibrating in its 2nd harmonic (2 loops). The number of nodes is n + 1 = 2 + 1 = 3 nodes (at x = 0, 0.6 m, 1.2 m). The speed of the wave is v = ω / k = (250π) / (5π / 3) = 150 m/s.

medium

4. Two identical strings of the same material, radius, and length are tuned to unison at 600 Hz. If the tension in one string is increased by 2.01%, the number of beats produced per second when they are sounded together is approximately:

  1. 3
  2. 9
  3. 12
  4. 6
Show answer and explanation

Answer: (D) 6

Since f ∝ √T, for small fractional changes, Δf / f ≈ (1/2) (ΔT / T). Here, ΔT / T = 0.0201. Therefore, Δf ≈ (1/2) × 0.0201 × 600 Hz = 6.03 ≈ 6 beats per second.

easy

5. Two sound waves are represented by y₁ = A sin(100πt) and y₂ = A sin(108πt), where t is in seconds. The beat frequency and the time interval between two successive maxima of sound intensity are, respectively:

  1. 8 Hz and 0.125 s
  2. 8 Hz and 0.25 s
  3. 4 Hz and 0.5 s
  4. 4 Hz and 0.25 s
Show answer and explanation

Answer: (D) 4 Hz and 0.25 s

The angular frequencies are ω₁ = 100π rad/s and ω₂ = 108π rad/s. The corresponding frequencies are f₁ = ω₁ / (2π) = 50 Hz and f₂ = ω₂ / (2π) = 54 Hz. The beat frequency is f_beat = |f₂ - f₁| = 54 - 50 = 4 Hz. The time interval between successive maxima is Δt = 1 / f_beat = 1 / 4 s = 0.25 s.

hard

6. A tuning fork produces 4 beats per second with an open organ pipe of length 50 cm vibrating in its fundamental mode. The same tuning fork produces 4 beats per second with another open organ pipe of length 51 cm vibrating in its fundamental mode. Assuming no end correction, the speed of sound in air is:

  1. 340 m/s
  2. 320 m/s
  3. 408 m/s
  4. 360 m/s
Show answer and explanation

Answer: (C) 408 m/s

Let f be the frequency of the fork. For pipe 1, f₁ = v / (2 × 0.50) = v / 1.0 = v. For pipe 2, f₂ = v / (2 × 0.51) = v / 1.02. Since length 50 cm is shorter, f₁ > f₂. The fork frequency f lies between f₁ and f₂: f₁ - f = 4 and f - f₂ = 4. Adding these gives f₁ - f₂ = 8. Thus, v - v / 1.02 = 8, which means v(1 - 1/1.02) = 8 => v(0.02 / 1.02) = 8 => v = 8 × 1.02 / 0.02 = 8 × 51 = 408 m/s.

hard

7. An organ pipe closed at one end vibrates in its fundamental mode with a frequency of 200 Hz. If the speed of sound in air is 340 m/s and an end correction of 0.6 cm is taken into account at the open end, the actual length of the pipe is:

  1. 84.4 cm
  2. 42.5 cm
  3. 43.1 cm
  4. 41.9 cm
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Answer: (D) 41.9 cm

For a closed pipe in fundamental mode, effective length L' = L + e = v / (4f). Here, L' = 340 / (4 × 200) = 340 / 800 = 0.425 m = 42.5 cm. Since e = 0.6 cm, the actual length of the pipe is L = L' - e = 42.5 cm - 0.6 cm = 41.9 cm.

hard

8. Tuning fork A has frequency 384 Hz. Sounding A with fork B gives 4 beats per second. When B is loaded with a little wax, the number of beats decreases. The original frequency of B is:

  1. 376 Hz
  2. 392 Hz
  3. 388 Hz
  4. 380 Hz
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Answer: (C) 388 Hz

B is 384 ± 4 Hz. Loading lowers B's frequency; beats decrease only if B was above A, so B = 384 + 4 = 388 Hz.

hard

9. Tuning fork A has frequency 384 Hz. Sounding A with fork B gives 7 beats per second. When B is loaded with a little wax, the number of beats decreases. The original frequency of B is:

  1. 370 Hz
  2. 377 Hz
  3. 398 Hz
  4. 391 Hz
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Answer: (D) 391 Hz

B is 384 ± 7 Hz. Loading lowers B's frequency; beats decrease only if B was above A, so B = 384 + 7 = 391 Hz.

hard

10. Tuning fork A has frequency 256 Hz. Sounding A with fork B gives 3 beats per second. When B is loaded with a little wax, the number of beats decreases. The original frequency of B is:

  1. 262 Hz
  2. 259 Hz
  3. 253 Hz
  4. 250 Hz
Show answer and explanation

Answer: (B) 259 Hz

B is 256 ± 3 Hz. Loading lowers B's frequency; beats decrease only if B was above A, so B = 256 + 3 = 259 Hz.

easy

11. In a pipe of length L closed at one end, which of the following statements correctly describes the displacement and pressure conditions at the closed end in any normal mode of vibration?

  1. Displacement antinode and pressure antinode
  2. Displacement node and pressure node
  3. Displacement node and pressure antinode
  4. Displacement antinode and pressure node
Show answer and explanation

Answer: (C) Displacement node and pressure antinode

At the rigid closed end of a pipe, air particles are prevented from moving, which creates a displacement node. Because air molecules pile up against this fixed boundary, the variation in pressure is maximum, forming a pressure antinode.

hard

12. Tuning fork A has frequency 480 Hz. Sounding A with fork B gives 7 beats per second. When B is loaded with a little wax, the number of beats decreases. The original frequency of B is:

  1. 466 Hz
  2. 473 Hz
  3. 487 Hz
  4. 494 Hz
Show answer and explanation

Answer: (C) 487 Hz

B is 480 ± 7 Hz. Loading lowers B's frequency; beats decrease only if B was above A, so B = 480 + 7 = 487 Hz.

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